Answer:
25.97oC
Explanation:
Heat lost by aluminum = heat gained by water
M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]
Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC
Let Temp(Al+H2O) = X
23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)
21.15(65.9-X) = 230.23(X-22.3)
1393.785 - 21.15X = 230.23X – 5134.129
230.23X + 21.15X = 1393.785 + 5134.129
251.38X = 6527.909
X = 6527.909/251.38
X = 25.97oC
So, the final temperature of the water and aluminum is = 25.97oC
The reactants that would lead to a spontaneous reaction
are Mn2+ and Al3+. Al and Mn
are not because they are stable reactants. Mn and Al3+ are not because Mn is stable. Mn2+ and Al are
not because Al is stable.
<h2><u>
Answer:</u></h2>
(These are not rounded to the correct decimal)
130.94 atm
13,266.6 kPa
99,571.4 mmHg
<h2><u>
Explanation:</u></h2>
<u></u>
PV = nRT
V = 245L
P = ?
R = 0.08206 (atm) , 8.314 (kPa) , 62.4 (mmHg)
T = 273.15 + 27 = 300.15K
n = 1302.5 moles
How I found (n).
5.21kg x 1000g/1kg x 1 mole/4.0g = 1302.5 moles
Now, plug all the numbers into the equation.
Pressure in atm = (1302.5)(0.08206)(300.15) / 245 = 130.94 atm (not rounded to the correct decimal)
Pressure in kPa = (1302.5)(8.314)(300.15) / 245 = 13,266.6 kPa (not rounded to the correct decimal)
Pressure in mmHg = (1302.5)(62.4)(300.15) / 245 = 99,571.4 mmHg (not rounded to the correct decimal)
